Question: A circle has a radius of $6$. An arc in this circle has a central angle of $40^\circ$. What is the length of the arc? ${12\pi}$ ${40^\circ}$ $\color{#DF0030}{\dfrac{4}{3}\pi}$ ${6}$
Solution: First, calculate the circumference of the circle. $c = 2\pi r = 2\pi (6) = 12\pi$ The ratio between the arc's central angle $\theta$ and $360^\circ$ is equal to the ratio between the arc length $s$ and the circle's circumference $c$ $\dfrac{\theta}{360^\circ} = \dfrac{s}{c}$ $\dfrac{40^\circ}{360^\circ} = \dfrac{s}{12\pi}$ $\dfrac{1}{9} = \dfrac{s}{12\pi}$ $\dfrac{1}{9} \times 12\pi = s$ $\dfrac{4}{3}\pi = s$